[hihi123]

Here's the review question:

S + HNO3 ----> SO2 + NO + H2O

a) write the oxidation half reaction
b)write the reduction half reaction
c) write the balanced equation for the above reaction

I am having trouble assigning oxidation numbers, for I don't know whether to count NO3 as a polyatomic or an two separate elements

[Donaldson Tan]

consider HNO3 as H⁺ and NO₃^-

your oxidation reaction is: S -> SO₂

whereas your reduction reaction is: NO₃^- -> NO

hope this helps in figuring out the mole/material balance

[hihi123]

So how am I suppossed to figure out which element is reduced  if I dont split up NO3? Ah. Usually I can get these but this one is really confusing me.

[Donaldson Tan]

HNO3 is a strong mineral acid. it dissociates readily to yield those ions.

HNO3 -> H+ + NO3-

It takes some effort to read up and remember all this little intrinsic details during your course of study.

[hihi123]

I understand that, but what im having trouble with is when I assign the oxidation numbers, I don't know if I should put a number for N and the O3. I know that S oxidizes, but i can't figure out the reduction

[Garneck]

I understand that, but what im having trouble with is when I assign the oxidation numbers, I don't know if I should put a number for N and the O3. I know that S oxidizes, but i can't figure out the reduction

Ok. What's the oxiadtion state of oxygen? -2, right?

So you can count the oxidation state of nitrogen in NO₃^- in a simple equation:

N = x
O = -2

x + 3*(-2) = -1 (this is the charge of NO₃^-)
x -6 = -1
x = 5

so nitrogen is on +5

[jdurg]

Another little hint here is that the hydrogen and oxygen atoms in this reaction do not change their oxidation number.  Hydrogen remains as +1 and Oxygen stays at -2.  So if you just focus on the atoms that do change in oxidation number (Sulfur and Nitrogen), you'll be able to easily write out those half reactions.  Once you've written out the half reactions, you just balance those so that the number of electrons is equal.  This will then tell you how many sulfur atoms you need in the balanced equation and how many nitrogen atoms you will need.  From there, it becomes nice and easy.  

[xiankai]

in my school, i learned that oxidation and reduction can be defined in 4 types, gain/loss of hydrogen/oxygen/electron/oxidation number.

the gain of oxygen is supposed to be oxidation, while the loss is reduction. this is an easyn way to figure it out, without going into too much depth. but its just only the basic concept.

[Donaldson Tan]

Oxidation is defined as gain of oxygen, or loss of electrons
Reduction is defined as loss of oxygen, or gain of electrons

Incorporating these 3 definitions, here are the three half equations to describe your reaction:

1. NO₃^- => NO + 2[

• ] + e

2. S + 2[

• ] => SO₂
  
  3. 2H⁺ + [
• ] + 2e => H₂O
  
  to combine and balance these 3 equations, you will need 4 sets of (1), 3 set of (2) and 2 set of (3). ie.
  
  4H⁺ + 3S + 8[
• ] +  4NO₃^- + 4e => 4NO + 8[
• ] + 4e + 3SO₂ + 2H₂O
  
  Cancelling the common terms on both reactant and product side, the net reaction is:
  3S + 4HNO₃ => 4NO + 3SO₂ + 2H₂O